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I'm looking for an example of a map $f : X \to Y$, where $X$ and $Y$ are manifolds (without boundary), and $f$ is a Serre fibration, but $f$ is not a fiber bundle.

I know that if $f$ is proper, and $f$ is a smooth fibration, then there are no such examples (every such proper submersion is a fiber bundle by Ehressman's theorem, and we lift smooth paths defining tangent vectors to prove that $f$ is a submersion.) All the examples of Serre fibrations of manifolds that I know (which are basically all principle bundles coming from Lie groups and other associated constructions) are fiber bundles. The other examples of fibrations I know are probably not fiber bundles but are not maps between manifolds (such as the path loop fibration).

Elle Najt
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  • How about a disk mapping to an interval (either with or without boundary) by projection? – Shay Ben-Moshe Mar 10 '18 at 16:13
  • @ShayBenMoshe For the boundaryless case it is a fiber bundle, though. For the boundary case -- that's a good example, thanks. Let's say I'm interesting in manifolds without boundary though. – Elle Najt Mar 14 '18 at 04:31
  • Yeah, you are right. I don't have an example without boundary :/ – Shay Ben-Moshe Mar 14 '18 at 15:30
  • I do have the feeling that projections won't work, because the trick of squeezing a line to a point doesn't seem to work, but I'm not sure even about that. – Shay Ben-Moshe Mar 14 '18 at 15:34
  • @Lorenzo here's an MO link with some added thoughts in the body of the question. Seems an example would be rather unpleasant. – Arrow May 10 '19 at 22:38
  • @Arrow What about $(x,y) \to (xy)^2$ on the punctured plane $\mathbb{R}^2 \setminus 0$? Or $|xy|$ if you don't care about smoothness. I think it cannot be locally trivialized at zero, because you can find $a(t)$, $b(t)$ on different components of the fiber, such that they both converge to the same point as $t \to 0$. I'm not sure if this is a fibration, but lifting seems okay (as in your now deleted answer), and also all the fibers are homeomorphic to the disjoint union of four intervals. – Elle Najt May 11 '19 at 22:43
  • Here's a proof just for the sake of completeness. Suppose the bundle was trivial about zero, of the form $B\times F\to B$ where $F$ is the fiber. Suppose $F=F_1\amalg F_2$. Then $B\times F\cong (B\times F_1)\amalg (B\times F_2)$. Thus the subsets $B\times F_1,B\times F_2$ can't share a limit point. (It's cool that the crux is the distributivity of the category of topological spaces.) – Arrow May 12 '19 at 11:41
  • @Lorenzo meh. Useful for understanding the HLP though. – Arrow May 12 '19 at 19:38
  • @Lorenzo maybe you should post an answer with a link to the MO example, just so this question won't appear "unanswered". – Arrow May 19 '19 at 07:56

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